Hans has $27 which he decides to spend on x and y. Commodity x costs $16 per unit and commodity y costs $10 per unit. He has the utility function U(x, y)- 5x2+2y2 and he can purchase fractional units of x and y. Hans will choose:a. only X b. only yc. some of each commodity but more y than x. d. some of each commodity but more x than y.

Option C is correct.

Hans will choose some of each commodity but more y than x.

The number of units of x and y that will maximize Hans' utility and fit well with his budget constraint are 0.854 units of x and 1.334 units of y.

Step-by-step explanation:

Hans wants to maximize utility but has a cost constraint. It seems like this is a 'finding the maximum of a multivariable function' problem.

U(x,y) = 5x² + 2y²

The constraint can be given as

Sum of the Cost of the number units of each commodity he will get must not exceed $27

Price of one commodity x = $16

Price of one commodity y = $10

16x + 10y ≤ 27

But since fractional units are possible,

16x + 10y = 27.

So, we want to maximize U(x,y) = 5x² + 2y² subject to the function of 16x + 10y = 27.

There are a number of ways to go About this, but the best of them all is the Lagrange function method.

We will write a Lagrange function,

L(x,y) = U(x,y) - λ(constraint function)

Constraint function = 16x + 10y - 27

L(x,y) = 5x² + 2y² - λ(16x + 10y - 27)

where λ can be a function of x and y

So, we take the derivatives of L with respect to x, y and λ. The values of x and y at this point are our answers.

L(x,y) = 5x² + 2y² - λ(16x + 10y - 27)

(∂L/∂x) = 10x - 16λ

At the maximum value we require, (∂L/∂x) = (∂L/∂y) = (∂L/∂λ) = 0

10x - 16λ = 0

λ = 10x/16 = 5x/8

(∂L/∂y) = 4y - 10λ

4y - 10λ = 0

λ = 4y/10 = 2y/5

(∂L/∂λ) = 16x + 10y - 27

16x + 10y - 27 = 0

Equating the values of λ obtained from the first two partial derivatives

λ = (5x/8) = (2y/5)

x = 16y/25

Putting this value for x in the third partial derivatives or the constraint function

16x + 10y - 27 = 0

16(16y/25) + 10y = 27

(256y/25) + 10y = 27

506y/25 = 27

506y = 27×25

y = 27×25/506 = 675/506 = 1.334 units.

x = 16y/23 = 0.854 units.

So, the number of units of x and y that will maximize Hans' utility and fit well with his budget constraint are 0.854 units of x and 1.334 units of y.

this was already solved by me with a different person. if you look it up in the search bar at the top of the page, you will get the answer.

step-by-step explanation:

answer:  

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

which is the required equation for finding the value of x,